# Why the balloon does not pop when there are a lot of needles?

There is a famous experiment where a balloon pops when we push it against a needle, but it doesn't pop when we push it against a lot of needles. Why is this?

A link to that experiment will be welcome, but even without it :

If you push the balloon against a needle, this single needle will have to resist to your push on a very small area of the balloon, (its tip, essentially) thus exerting a very high pressure that would burst the skin of the balloon before the general shape of the balloon changes enough to adapt.

If you push it against many needles, the push will be shared and the pressure on each tip will be divided by the number of needles. And if you try to push harder, it will not work so easily, because then the whole balloon will change shape and spread in directions where there are no needles, without exerting much more push on the general area where the set of needles is concentrated.

• Very much like lying down on a bed of nails - if there's one nail, it's going to go right through you. Many many nails? A nice, comfy place to rest. (Okay, maybe not comfy....) Jan 4 at 16:48
• @BruceWayne Just so ! Jan 4 at 16:50
• Am I the only one delighted by an exchange between BruceWayne and Alfred? Jan 4 at 19:08
• Consider the case of a bed of infinitely thin needles all next to each other. Any one of them individually is sharp, but all together, it’s just a solid sheet of metal. Jan 5 at 0:22
• @fyrepenguin Sure, but you dont need to go that far. Even when the needles are well apart, by much more than the width of each needle, if the number of needles is large enough the balloon will not pop. Jan 5 at 0:49

Many needles minimizes pressure, given same pushing force as per : $$p= \frac F A = \frac {F}{kA_0}$$ where $$A_0$$ is needle tip area and $$k$$ - number of needles. So as number of needles $$k$$ increases - "pushing pressure" decreases, given same pushing force $$F$$.